14 Diagonalization 14.1 Change of basis 2 Let = (b1, b2) be an ordered basis for R and let B = b1 b2 (the matrix B 2 with b1 and b2 as columns). If x is a vector in R , then its coordinate vector [x]B relative to satisfies the formula B B[x]B = x T We can see this by writing [x]B = [α1, α2] to get α1 B[x]B = b1 b2 = α1b1 + α2b2 = x. α2 The columns of B are linearly independent (since they form a basis) so B has full rank and is therefore invertible. This allows us to solve the equation above for the coordinate vector: −1 [x]B = B x 2 14.1.1 Example Let = (b1, b2) be the ordered basis of R with b1 = T T B [1, 2] and b2 = [2, 3] . Use the formula above to find the coordinate vector T [x]B where x = [1, 1] . − Solution We have 1 2 B = b1 b2 = , 2 3 so −1 1 3 2 1 5 5 [x]B = B x = − = = − . 1 2 1 1 − 3 3 − − − − 1 2 1 (Check: ( 5)b1 + (3)b2 = 5 +3 = = x.) − − 2 3 1 − 2 Let = (c1, c2) be another ordered basis for R and put C = c1 c2 . For anyC vector x in R2, the coordinate vectors of x relative to and satisfy the equation B C B[x]B = C[x]C 1 14 DIAGONALIZATION 2 This equation holds since both sides equal x by the first equation of the section. We can solve for [x]B to get −1 [x]B = B C[x]C P | {z } The matrix P = B−1C is called the “transformation matrix from to .” C B 14.1.2 Example Let = (b1, b2) and = (c1, c2) be the ordered bases of R2 with B C 2 1 0 1 b1 = , b2 = , c1 = , c2 = . 6 4 1 5 − and let x and y be vectors in R2. (a) Find the transformation matrix P from to . C B T (b) Given that [x]C = [ 3, 6] , find [x]B. − T (c) Given that [y]B = [7, 2] , find [y]C . Solution (a) We have −1 −1 2 1 0 1 1 4 1 0 1 1 1 1 P = B C = = − = − 6 4 1 5 2 6 2 1 5 2 2 4 − − − − 1 1 2 2 = − . 1 2 "− # (b) We have 1 1 9 2 2 3 2 [x]B = P[x]C = − − = − . 1 2 6 15 "− # " # (c) Multiplying both sides of the change of basis formula by P−1 we get 1 −1 1 2 2 7 15 30 [y]C = P [y]B = =2 = . 1 1 1 2 8 16 2 " 2 # Here is the general formulation for a change of basis: 14 DIAGONALIZATION 3 Change of basis. Let = (b1, b2,..., bn) and = (c1, c2,..., cn) be two or- B n C dered bases for R , and put B = b1 b2 bn and C = n ··· c1 c2 cn . For any vector x in R , we have ··· B[x]B = C[x]C so that −1 [x]B = B C[x]C. P −1 The matrix P = B C is called| {z the} transformation matrix from to . C B 14.2 Linear function and basis change Theorem. Let L : Rn Rn be a linear function, let and be ordered bases for Rn→, let P be the transformation matrixB fromC to , and let A be the matrix of L relative to . The matrixC of BL relative to is P−1AP, that is, B C −1 [L(x)]C = P AP[x]C for all x Rn. ∈ Proof. For all x Rn we have ∈ −1 −1 P AP[x]C = P A[x]B definition of transformation matrix −1 = P [L(x)]B A is matrix of L relative to −1 B = [L(x)]C P is transformation matrix from to . B C 14.2.1 Example Let L : R2 R2 be the linear function given by → 1 3 2 x1 + 2 x2 L(x)= 3 1 . " 2 x1 + 2 x2# (a) Find the matrix A of L (relative to the standard ordered basis = E (e1, e2)). 14 DIAGONALIZATION 4 (b) Find the transformation matrix P from = ([1, 1]T , [ 1, 1]T ) to . C − E (c) Use the theorem to find the matrix of L relative to . C Solution (a) We have 1 3 2 2 A = L(e1) L(e2) = 3 1 . " 2 2 # (b) Writing E = e1 e2 we see that E = I, so that −1 −1 1 1 P = E C = I C = C = . 1− 1 (c) According to the theorem, the matrix of L relative to is C −1 1 3 −1 −1 1 1 2 2 1 1 P AP = C AC = 1− 1 3 1 1− 1 " 2 2 # 1 3 1 1 1 2 2 1 1 = − 2 1 1 3 1 1 1 − " 2 2 # 1 2 2 1 1 = − 2 1 1 1 1 − 1 4 0 2 0 = = 2 0 2 0 1 − − In this example, the matrix of L relative to the basis turned out to be much simpler than the matrix relative to the standard basis.C The reason is that is a basis for R2 consisting of eigenvectors of L. We investigate such bases inC the next section. 14.3 Method for diagonalization The matrix 2 0 0 0 3 0 0− 0 1 is an example of a “diagonal matrix.” In general, a diagonal matrix is a matrix having the property that every entry not on the main diagonal is 0. An n n matrix A is diagonalizable if there exists an invertible n n matrix P such× that P−1AP = D, where D is a diagonal matrix. × 14 DIAGONALIZATION 5 Theorem. Let A be an n n matrix. The matrix A is diagonalizable if and only if there× exists a basis for Rn consisting of eigenvectors of A. In this case, if P is the matrix with the eigenvectors as columns, then P−1AP = D with D diagonal. Proof. We will prove only the case n = 2. Assume that there exists a basis 2 b1, b2 for R consisting of eigenvectors of A. Let λ1 and λ2 be the corre- { } λ1 0 sponding eigenvalues and write D = . We have 0 λ2 λ1 0 AP = A b1 b2 = Ab1 Ab2 = λ1b1 λ2b2 = b1 b2 = PD 0 λ2 Therefore, P−1AP = D. Conversely, if P diagonalizes A, then the equation above shows that the columns of P must be eigenvectors of A and these columns are linearly independent since P is invertible. 1 1 14.3.1 Example Let A = . If possible, find an invertible 2 2 2 4 × −1 − matrix P such that P AP = D, where D is a diagonal matrix. Solution According to the theorem, such a matrix P exists if and only if there exists a basis for R2 consisting of eigenvectors of A. The characteristic polyno- mial of A is 1 λ 1 2 det(A λI)= − = (1 λ)(4 λ)+2 = λ 5λ+6=(λ 2)(λ 3). − 2 4 λ − − − − − − − The eigenvalues of A are the zeros of this polynomial, namely, λ =2, 3. Next, the λ-eigenspace of A is the solution set of the equation (A λI)x = 0: − (λ = 2) 1 1 0 1 1 0 [A 2I 0] = − − , − | 2 2 0 ∼ 0 0 0 − so the 2-eigenspace of A is [t,t]T t R . Letting t = 1, we get a 2-eigenvector [1, 1]T ; { | ∈ } 14 DIAGONALIZATION 6 (λ = 3) 2 1 0 1 1 0 [A 3I 0] = − − 2 , − | 2 1 0 ∼ 0 0 0 − t T so the 3-eigenspace of A is [ 2 ,t] t R . Letting t = 2 (to avoid fractions), we get a 3-eigenvector{ [1, 2]| T .∈ } The eigenvectors [1, 1]T and [1, 2]T of A form a basis for R2 (neither is a multiple of the other so they are linearly independent; since dim R2 = 2, they form a basis). According to the theorem, the matrix P with these vectors as columns should have the indicated property: 1 1 P = . 1 2 Computing we get −1 −1 1 1 1 1 1 1 1 2 1 1 1 1 1 P AP = = − 1 2 2 4 1 2 1 1 1 2 4 1 2 − − − 4 2 1 1 2 0 = = = D. 3− 3 1 2 0 3 − (Note that the eigenvalues of A appear along the main diagonal of D.) 0 1 14.3.2 Example Is the matrix A = diagonalizable? Explain. 1− 0 Solution The characteristic polynomial of A is λ 1 2 det(A λI)= − − = λ +1. − 1 λ − Since this polynomial has no zeros (in R), the matrix A has no eigenvalues. Therefore, there is no basis for R2 consisting of eigenvectors of A and A is not diagonalizable according to the theorem. (Here’s another way to see that A has no eigenvalues. A is the matrix of the linear function “90◦ clockwise rotation.” Since this function sends no nonzero vector to a multiple of itself, it has no eigenvalues, and therefore A has no eigenvalues either.) 14.4 Power of matrix It is often necessary to compute high powers of a square matrix A, such as A10. Just multiplying the matrix A by itself over and over again can be quite tedious. 14 DIAGONALIZATION 7 However, if A is diagonalizable, then there is an observation that greatly reduces the number of computations: Let A be a diagonalizable matrix, so that P−1AP = D with D diagonal.